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给你一棵以 root 为根的二叉树和一个 head 为第一个节点的链表。如果在二叉树中，存在一条一直向下的路径，且每个点的数值恰好一一对应以 head ,"> 
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        <h1 class="title">leetcode5346. 二叉树中的列表</h1>
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            <span>三月 14, 2020</span>
            

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            <h1 id="leetcode5346-二叉树中的列表"><a href="#leetcode5346-二叉树中的列表" class="headerlink" title="leetcode5346. 二叉树中的列表"></a>leetcode5346. 二叉树中的列表</h1><hr>
<blockquote>
<p>给你一棵以 root 为根的二叉树和一个 head 为第一个节点的链表。<br>如果在二叉树中，存在一条一直向下的路径，且每个点的数值恰好一一对应以 head 为首的链表中每个节点的值，那么请你返回 True ，否则返回 False 。<br>一直向下的路径的意思是：从树中某个节点开始，一直连续向下的路径。</p>
</blockquote>
<blockquote>
<p>示例 1：</p>
</blockquote>
<p><img src="https://imgconvert.csdnimg.cn/aHR0cHM6Ly9hc3NldHMubGVldGNvZGUtY24uY29tL2FsaXl1bi1sYy11cGxvYWQvdXBsb2Fkcy8yMDIwLzAyLzI5L3NhbXBsZV8xXzE3MjAucG5n?x-oss-process=image/format,png" alt=""></p>
<blockquote>
<p>输入：head = [4,2,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]<br>输出：true<br>解释：树中蓝色的节点构成了与链表对应的子路径。</p>
</blockquote>
<p>第一次打周赛，死的很惨，看了一下评论区的解法，我改进了一下我当时的提交。开始我想写一个方法，既能遍历所有的节点，又能查找子串，但是发现这么写难度很大，而且好不容易改好一个提交还超时。<br>对于最后一组样例，我觉得可能跟这个类似：</p>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">[2,2,1]</span><br><span class="line">[1,1,2,null,null,null,2,2,2,1,2,1,2]</span><br></pre></td></tr></table></figure>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br></pre></td><td class="code"><pre><span class="line"><span class="class"><span class="keyword">class</span> <span class="title">Solution</span> </span>&#123;</span><br><span class="line">    <span class="keyword">public</span> <span class="keyword">boolean</span> flag=<span class="keyword">false</span>;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">boolean</span> <span class="title">isSubPath</span><span class="params">(ListNode head, TreeNode root)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(root==<span class="keyword">null</span>)<span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line">        <span class="keyword">if</span>(head==<span class="keyword">null</span>)<span class="keyword">return</span> <span class="keyword">false</span>;</span><br><span class="line">        flag=<span class="keyword">false</span>;</span><br><span class="line">        tt(root,head);</span><br><span class="line">        <span class="keyword">return</span> flag;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="function"><span class="keyword">public</span> <span class="keyword">void</span> <span class="title">tt</span><span class="params">(TreeNode root,ListNode head)</span></span>&#123;</span><br><span class="line">    <span class="comment">//用来遍历每个节点</span></span><br><span class="line">        <span class="keyword">if</span>(root==<span class="keyword">null</span>)<span class="keyword">return</span>; </span><br><span class="line">        <span class="keyword">if</span>(!flag)&#123;</span><br><span class="line">            traverse(root,head);</span><br><span class="line">            <span class="keyword">if</span>(flag==<span class="keyword">false</span>)&#123;</span><br><span class="line">                tt(root.left,head);</span><br><span class="line">                tt(root.right,head);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125; </span><br><span class="line">   <span class="function"><span class="keyword">public</span> <span class="keyword">void</span> <span class="title">traverse</span><span class="params">(TreeNode root,ListNode head)</span></span>&#123;</span><br><span class="line">   <span class="comment">//从当前节点开始查找字符串</span></span><br><span class="line">       <span class="keyword">if</span>(head==<span class="keyword">null</span>)&#123;</span><br><span class="line">           flag=<span class="keyword">true</span>;</span><br><span class="line">           <span class="keyword">return</span>;</span><br><span class="line">       &#125;</span><br><span class="line">       <span class="keyword">if</span>(root==<span class="keyword">null</span>)<span class="keyword">return</span>;</span><br><span class="line">       <span class="keyword">if</span>(root.val==head.val)&#123;</span><br><span class="line">       traverse(root.left,head.next);</span><br><span class="line">       traverse(root.right,head.next);</span><br><span class="line">       &#125;</span><br><span class="line">   &#125;   </span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<p><strong>leetcode 43/100</strong></p>

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